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k0ka [10]
4 years ago
12

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wi

re. The company you work for is poorly equipped. You find a battery, a voltmeter, and an ammeter, but no meter for directly measuring resistance (an ohmmeter). You put the leads from the voltmeter across the terminals of the battery, and the meter reads 12.1 . You cut off a 20.0- length of wire and connect it to the battery, with an ammeter in series with it to measure the current in the wire. The ammeter reads 6.50 . You then cut off a 40.0- length of wire and connect it to the battery, again with the ammeter in series to measure the current. The ammeter reads 4.50 . Even though the equipment you have available to you is limited, your boss assures you of its high quality: The ammeter has very small resistance, and the voltmeter has very large resistance.
What is the resistance of 1 meter of wire?

Physics
1 answer:
krek1111 [17]4 years ago
6 0

Explanation:

Below is an attachment containing the solution.

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8 0
3 years ago
Water leaks out of a 3,200-gallon storage tank (initially full) at the rate V '(t) = 80 -t, where t is measured in hours and V i
kvasek [131]

Answer:

water leak is 650 gallons

time required to full drain is 80 hrs

Explanation:

given data

volume V = 3200 gallon

rate = V(t) = 80 - t

to find out

how much water leak between 10 and 20 hour and  drain complete

solution

we know here rate is 80 - t

so here rate will be

\frac{dV(t)}{dt} = 80 - t

and for time 10 and 20 hour

take integrate between 10 and 20

so water leak = \int\limits^ {20}_ {10} {(80-t)} \, dt   .....................1

water leak = ( 80t - \frac{t^{2} }{2} )^{20}_{10}

water leak = 650

so water leak is 650 gallons

and

we know here for full tank drain condition

water leak full = 80 t - \frac{t^{2} }{2}

3200  = 80 t - \frac{t^{2} }{2}

6400 = t² - 160 t

t = 80

so time required to full drain is 80 hrs

6 0
4 years ago
The mass of an object is determined by: weighing the object with a scale multiplying the height, width, and length measuring the
rosijanka [135]
The true scientific way is the last: using the water displacement method
7 0
3 years ago
Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o
sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

a_{x} = -\omega^{2}  x

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders  is given as:

                                  I=\frac{1}{2}MR^{2} -----(1)

Newton's second law for angular motion is:

                                             ∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

                                             ∑Fₓ = Maₓ ------ (3)

By definition of torque:

                                               τ  = RF --------(4)        

Put (4) and (1) in (2)

                                       RF=\frac{1}{2}MR^{2}\alpha

                                       RF=\frac{1}{2}MR^{2}\alpha

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

                                   f_{s}=\frac{1}{2}MR\alpha------ (5)

As angular acceleration is related to linear by:

                                          a= R\alpha

Eq (5) becomes

                                    f_{s}=\frac{1}{2}Ma_{x}---- (6)

aₓ shows displacement in horizontal direction

From (3)

                                              ∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

                                  \sum F_{x} = f_{s} - kx ----(7)

Put (7) in (3)

                                  f_{s} - kx  = Ma_{x}[/tex] -----(8)

Using (6) in (8)

                               \frac{1}{2}Ma_{x} - kx =Ma_{x}

                                     a_{x} = \frac{2k}{3M} x --- (9)

For spring mass system

                                  a= -\omega^{2} x ----- (10)

Equating (9) and (10)

                                  \omega^{2} = \frac{2k}{3M}

\omega = \sqrt{ \frac{2k}{3M}}

then (9) becomes

                                a_{x} = - \omega^{2}x

(The minus sign says that x and  aₓ  have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

                                   T=\frac{2\pi }{\omega}

                                  T = 2\pi \sqrt{\frac{3M}{2k}

                           

 

5 0
3 years ago
A rear window defroster consists of a long, flat wire bonded to the inside surface of the window. When current passes through th
GenaCL600 [577]

Answer:

\rho=1.8\times 10^{-8}\Omega.m

Explanation:

Given:

width of the wire, w=1.8\ mm=1.8\times 10^{-3}\ m

thickness of the flat wire, d=0.12\ mm=1.2\times 10^{-4}

length of the wire, l=12\ m

voltage across the wire, V=12\ V

current through the wire, I=12\ A

Now the net resistance of the wire:

using ohm's law

R=\frac{V}{I}

R=\frac{12}{12}

R=1\ \Omega

We have the relation between the resistivity and the resistance as:

R=\rho.\frac{l}{a}

where:

a = cross sectional area of the wire

\rho = resistivity of the wire material

1=\rho\times \frac{12}{1.8\times 10^{-3}\times 1.2\times 10^{-4}}

\rho=1.8\times 10^{-8}\Omega.m

4 0
4 years ago
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